Lecture 10 - Adjoint in More Detail

Recall the idea from last time:

This diagram gives the idea of the maps we are using, where $f (\vec{w}) = ⟨ \vec{w}, \vec{f} ⟩$ for all $\vec{w}$ .

We've spent a long time talking about the adjoint. As such, let's look at some nice properties of certain operators.

self-adjoint

An operator $T \in L (V)$ is called self-adjoint if $T^{*} = T$ :

⟨ T v_{1}, v_{2} ⟩ = ⟨ v_{1}, T v_{2} ⟩

for all $v_{1}, v_{2} \in V$ .

A lot of nice properties come as a result of this:

Lemma

The eigenvalues of a self-adjoint operator $λ$ are always real.

Proof
Suppose $T \vec{v} = λ \vec{v}$ for some non-zero $\vec{v} \in V$ . We'll show $λ$ has no imaginary component. Notice that:

\begin{aligned} ⟨ T v, v ⟩ & = ⟨ v, T v ⟩ \\ ⟨ λ v, v ⟩ & = ⟨ v, λ v ⟩ \\ λ v, v & = \overset{―}{λ} ⟨ v, v ⟩ \end{aligned}

so since $v \neq \vec{0}$ then $⟨ v, v ⟩ \neq 0$ so divide by it, showing $λ = \overset{―}{λ}$ so then $λ$ must be real.
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Properties of Adjoint Operators

$V$ is finite-dimensional complex inner product space.

$⟨ T v, v ⟩ = 0$ for all $v \in V$ is equivalent to $T = 0_{V}$ .
$⟨ T v, v ⟩ \in R$ for all $v \in V$ is equivalent to $T = T^{*}$ .

Proof
The proof of (a) comes from the idea that if $⟨ T u, v ⟩$ is a sum of things that look like $⟨ T *, * ⟩$ then the whole thing is 0 by assumption. The reverse direction is pretty obvious though.

For (b), notice if our supposition is true then:

\begin{aligned} \underset{= 0 by supposition}{\underset{⏟}{⟨ T v, v ⟩ - \overset{―}{⟨ T v, v ⟩}}} & = ⟨ T v, v ⟩ - ⟨ v, T v ⟩ \\ = ⟨ T v, v ⟩ - ⟨ T^{*} v, v ⟩ \\ = ⟨ (T - T^{*}) v, v ⟩ \end{aligned}

So then the whole thing is 0, so $T - T^{*}$ is the zero map, and thus must be equal (all iff).
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Proposition

If $T = T^{*}$ (ie: $T$ is self-adjoint) and $⟨ T v, v ⟩ = 0$ for all $v \in V$ , then $T = 0_{V}$ , considering $V$ is a real vector space

Proof
This is different from the previous lemma, as there we were in a complex inner product space. The proof is very similar, and we'll skip it for here.
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Normal Operators

normal operators

An operator $T \in L (V)$ is called normal if:

T T^{*} = T^{*} T

in other words, $T$ commutes with its adjoint.

Note that if $T = T^{*}$ , then this always happens. So all self-adjoint operators are normal. But the other way is not true.

As an example, consider $V = span (e_{1}, e_{2})$ with $e_{1}, e_{2}$ being a LI set. Define the map:

Γ \in L (V) : Γ (e_{1}) = e_{1} + 2 e_{2}, Γ (e_{2}) = - 2 e_{1} + e_{2}

thus:

M (Γ) = [\begin{matrix} 1 & - 2 \\ 2 & 1 \end{matrix}]

so then:

M (Γ^{*}) = [\begin{matrix} 1 & 2 \\ - 2 & 1 \end{matrix}] = M (Γ)^{*}

notice these don't equal, so $Γ$ is not self-adjoint. But is $Γ$ normal?

\begin{array}{r} M (Γ Γ^{*}) = M (Γ) M (Γ^{*}) = . . . = [\begin{array}{c} 5 & 0 \\ 0 & 5 \end{array}] \end{array}

and for the other way around:

M (Γ^{*} Γ) = [\begin{matrix} 5 & 0 \\ 0 & 5 \end{matrix}]

so $Γ$ is normal, while not being self-adjoint.

Nice properties of normal operators.

$T \in L (V)$ is normal operator iff $‖ T v ‖ = ‖ T^{*} v ‖$ for all $v \in V$ .

Proof
Start with $T$ is normal, iff $T T^{*} - T^{*} T = 0$ iff by Lecture 10 - Adjoint in More Detail#^64e249 then $⟨ (T T^{*} - T^{*} T) v, v ⟩ = 0$ for all $v \in V$ . Breaking up the right side:

\begin{aligned} ⟨ (T T^{*} - T^{*} T) v, v ⟩ = 0 & \Leftrightarrow ⟨ T T^{*} v, v ⟩ = ⟨ T^{*} T v, v ⟩ \\ \Leftrightarrow ⟨ T^{*} v, T^{*} v ⟩ = ⟨ T v, T v ⟩ & Def. of adjoint \\ \Leftrightarrow ‖ T^{*} v ‖^{2} = ‖ T v ‖^{2} \end{aligned}

Taking square roots finishes the proof.
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Eigenvectors for a normal operator

Suppose $T \in L (V)$ is normal, and a given eigenvector $v \in V$ for $T$ (ie: $T v = λ v$ where $v \neq \vec{0}$ ). Then $T^{*} v = \overset{―}{λ} v$ , so $v$ is an eigenvector of $T^{*}$ as well.

Proof
For (1), we know $(T - λ I)^{*} = T^{*} - (λ I)^{*} = T^{*} - \overset{―}{λ} I^{*} = T^{*} - \overset{―}{λ} I$ . As such, then we want to know if $T - λ I$ is a normal operator. For that to be the case then:

\begin{aligned} (T - λ I) (T - λ I)^{*} & = (T - λ I)^{*} (T - λ I) & ? \\ T T^{*} - λ T^{*} - \overset{―}{λ} T + | λ |^{2} & = T^{*} T - λ T^{*} - \overset{―}{λ} T + | λ |^{2} \end{aligned}

So clearly it's normal. So then next time we'll finish the proof.
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